Question

tìm x biết: \(\frac{6}{x^2+2}+\frac{12}{x^2+8}=3-\frac{7}{x^2+3}\)

Answer 1

Đặt \(x^2+3=t\) ta có:

\(\frac{6}{t-1}+\frac{12}{t+5}=3-\frac{7}{t}\)

\(\Leftrightarrow\frac{6\left(t+5\right)}{\left(t-1\right)\left(t+5\right)}+\frac{12\left(t-1\right)}{\left(t-1\right)\left(t+5\right)}=\frac{3t-7}{t}\)

\(\Leftrightarrow\frac{18t+18}{t^2+4t-5}=\frac{3t-7}{t}\)

\(\Leftrightarrow3t^3+5t^2-43t+35=18t^2+18t\)

\(\Leftrightarrow3t^3-13t^2-61t+35=0\)

\(\Leftrightarrow\left(t-7\right)\left(3t^2+8t-5\right)=0\)

\(\Leftrightarrow\left[\begin{matrix}t=7\\t=\frac{-8\pm\sqrt{124}}{6}\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x^2+3=7\\x^2+3=\frac{-8\pm\sqrt{124}}{6}\end{matrix}\right.\)

\(\Leftrightarrow\left[\begin{matrix}x^2=4\\x^2+3=\frac{-8\pm\sqrt{124}}{6}\left(loai\right)\end{matrix}\right.\)\(\Leftrightarrow x=\pm2\)